Determination of lead and tin by stripping voltammetry ... - Metrohm
Determination of lead and tin by stripping voltammetry ... - Metrohm Determination of lead and tin by stripping voltammetry ... - Metrohm
Application Bulletin 176/4 eDetermination of lead and tin by anodic stripping voltammetryPage 6/65 0STIR 20.06 SWEEP 25.6 U.start -800 mV U.step 4 mVU.end -300 mV Sweep rate 20 mV/s7 0MEAS U.standby mV8 ENDMethod: AB176_1bCALCULATIONmax. 15 lines--------------------------------------------------------------------------------Quantity Formula (R##, C##, A##) Res.unit Sig.dig.------------ ---------------------------------------- -------- --------Sn R1000=MC:Sn #g/L 5Pb R1001=MC:Pb #g/L 5Method 2: Determination of lead and tin withmethylene bluePrincipleTraces of Sn and Pb can be determined in oxalatebuffer in presence of methylene blue. Interferences byCd or Tl can be eliminated by modifying the pH andan intermediate electrolysis procedure.The detection limit of Sn and of Pb is 1 µg/L.ReagentsAll of the used reagents must be of purest qualitypossible (analytical grade or suprapur). Only ultrapurewater should be used.• Hydrochloric acid, suprapur, w(HCl) = 30%• Di-ammonium oxalate monohydrate, puriss. p. a.,CAS 6009-70-7• Ammonium chloride, suprapur, CAS 12125-02-9• Methylene blue, C.I. 52015, CAS 61-73-4• Sn standard stock solution, β(Sn 4+ ) = 1 g/L,commercially available• Pb standard stock solution, β(Pb 2+ ) = 1 g/L,commercially availableReady-to-use solutions:Base electrolytemethylene bluesolutionPb standardsolutionc(oxalate) = 0.14 mol/Lc(Ammoniumchlorid) = 0.17mol/Lc(HCl) = 0.15 mol/LpH = 1.619.2 g ammonium oxalate and9.2 g ammonium chloride aredissolved in ultrapure water.15.8 mL hydrochloric acid areadded. The solution is made upto 500 mL with ultrapure water.β(methylene blue) = 1g/L0.1 g methylene blue are dissolvedin 100 mL ultrapure water.The solution is stable for oneweek.β(Pb 2+ ) = 1 mg/LSn standardsolutionAnalysisβ(Sn 4+ ) = 1 mg/LThe solutions are diluted withc(HCl) = 0.01 mol/L. They arestable for max. 1 week.5 mL (diluted) sample+ 5 mL base electrolyte+ 0.05 mL methylene blue solutionThe pH of the solution should be 1.8.The voltammogram is recorded with the followingparameters:Working electrodeHMDEStirrer speed2000 rpmModeDPdrop size 4Purge time300 sCleaning potential (deposition) -800 mVCleaning time90 sDeposition potential-580 mV(intermediate electrolysis)Deposition timeEquilibration timePulse amplitudeStart potentialEnd potentialVoltage stepVoltage step timeSweep ratePeak potential (Sn)Peak potential (Pb)20 s10 s50 mV-800 mV-250 mV4 mV0.2 s20 mV/s-540 mV-400 mVThe concentration is determined by standard addition.
Application Bulletin 176/4 eDetermination of lead and tin by anodic stripping voltammetryPage 7/7ExampleDetermination of Pb and SnSubstance: Sn VR(**)I (nA)I (nA)I (nA)15105-600 -550 -500 -450U (mV)Standard addition curve: Sn12.5107.552.505 10 15 20 25 30 35rho(eff) (µg/L)Substance: Pb VR(**)15105Remarks• If the tin excess is great, one must work with twosegments, intermediate electrolysis (intermediateelectrolysis potential approx. -540 mV) and perhapstwo standard addition loops.• If the sample contains TI, the Pb peak can beadjusted to more positive values by raising the pHvalue to 2.4 (addition of ammonia solution w(NH 3)= 25%). One must work fast because at this pHvalue tin already hydrolyses.peak potentials:Pb-370 mVTl-410 mVSn-540 mVA good separation under these conditions can stillbe performed when the ratio Sn:TI lies at 1:2.Lead cannot be determined.• If the sample contains Cd, the pH value can belowered to 1.6 with hydrochlorid acid (w(HCI) =30%). The Sn peak adjusts then to more positivevalues improving the separation between Cd andSn. The Pb and the Tl concentration should, however,not be too high.peak potentials:Pb-400 mVSn-500 mVCd-600 mVFor separation, it is better to perform an intermediateelectrolysis(intermediate electrolysis potentialapprox. -580 mV). An excess Cd:Pb of 50:1does not show any interference.-500 -450 -400 -350U (mV)Standard addition curve: PbI (nA)1510501 2 3 4 5 6 7 8 9rho(eff) (µg/L)Sample VolumeResults5 mL27.4 µg/L Sn5.1 µg/L Pb
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Application Bulle<strong>tin</strong> 176/4 e<strong>Determination</strong> <strong>of</strong> <strong>lead</strong> <strong>and</strong> <strong>tin</strong> <strong>by</strong> anodic <strong>stripping</strong> <strong>voltammetry</strong>Page 7/7Example<strong>Determination</strong> <strong>of</strong> Pb <strong>and</strong> SnSubstance: Sn VR(**)I (nA)I (nA)I (nA)15105-600 -550 -500 -450U (mV)St<strong>and</strong>ard addition curve: Sn12.5107.552.505 10 15 20 25 30 35rho(eff) (µg/L)Substance: Pb VR(**)15105Remarks• If the <strong>tin</strong> excess is great, one must work with twosegments, intermediate electrolysis (intermediateelectrolysis potential approx. -540 mV) <strong>and</strong> perhapstwo st<strong>and</strong>ard addition loops.• If the sample contains TI, the Pb peak can beadjusted to more positive values <strong>by</strong> raising the pHvalue to 2.4 (addition <strong>of</strong> ammonia solution w(NH 3)= 25%). One must work fast because at this pHvalue <strong>tin</strong> already hydrolyses.peak potentials:Pb-370 mVTl-410 mVSn-540 mVA good separation under these conditions can stillbe performed when the ratio Sn:TI lies at 1:2.Lead cannot be determined.• If the sample contains Cd, the pH value can belowered to 1.6 with hydrochlorid acid (w(HCI) =30%). The Sn peak adjusts then to more positivevalues improving the separation between Cd <strong>and</strong>Sn. The Pb <strong>and</strong> the Tl concentration should, however,not be too high.peak potentials:Pb-400 mVSn-500 mVCd-600 mVFor separation, it is better to perform an intermediateelectrolysis(intermediate electrolysis potentialapprox. -580 mV). An excess Cd:Pb <strong>of</strong> 50:1does not show any interference.-500 -450 -400 -350U (mV)St<strong>and</strong>ard addition curve: PbI (nA)1510501 2 3 4 5 6 7 8 9rho(eff) (µg/L)Sample VolumeResults5 mL27.4 µg/L Sn5.1 µg/L Pb
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